Problem: $A=\left[\begin{array}{rr}10 & 2 & -8 & 9 \\3 & 6 & 1 & -4 \\-2 &1 &9 & 0 \\8 &-3 &-6 & 1 \\-6 &3 &9 & -1\end{array}\right]$ $A_{3,4}=$
Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{3,4}$ $A_{{3},{4}}$ is located on row ${3}$ of $A$ : $\left[\begin{array}{rr}10 & 2 & -8 & 9 \\3 & 6 & 1 & -4 \\{-2} & {1} & {9} & {0} \\8 &-3 &-6 & 1 \\-6 &3 &9 & -1\end{array}\right]$ $A_{{3},{4}}$ is also located on column ${4}$ of $A$. $\left[\begin{array}{rr}10 & 2 & -8 & 9 \\3 & 6 & 1 & {-4} \\{-2} & {1} & {9} & {\text0} \\8 &-3 &-6 & 1 \\-6 &3 &9 & {-1}\end{array}\right]$ Therefore, $A_{{3},{4}}={0}$. Summary $A_{3,4}=0$